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MathGroup Archive 2007

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Re: simplification of 0/0 to 1?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77605] Re: simplification of 0/0 to 1?
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Wed, 13 Jun 2007 07:47:31 -0400 (EDT)
  • References: <f4j11t$ksa$1@smc.vnet.net>

I have checked what two other CASs do in this simplification.
The one returns the expression unevaluated the other returns ?
which means undefined.
I guess hidden zero will be a problem somewhere else
for these CASs.

Because also I want to be fair with our lovely Mathematica
let me talk how this expression appeared.

 int(1-1/(sqrt(2)*(cos(Pi/12)-sin(Pi/12)))^z, z= 0..1);

  /           Pi                Pi            Pi            Pi
  |ln(2) cos(----) - ln(2) sin(----) - 2 cos(----) + 2 sin(----)
  \           12                12            12            12
            1/2             Pi               Pi  \   / /
         + 2    + 2 %1 cos(----) - 2 %1 sin(----)|  /  |
                            12               12  / /   \
                   Pi                Pi               Pi
        ln(2) cos(----) - ln(2) sin(----) + 2 %1 cos(----)
                   12                12               12
                     Pi  \
         - 2 %1 sin(----)|
                     12  /
                Pi          Pi
  %1 := ln(cos(----) - sin(----))

That is, another CAS returned a meshy expression
(the expression I talked about in the first message)
for this integral, while Mathematica (5.2)
returns 0 which is the correct result.

->So, Integrate does the job good.
->Simplify faces a somehow inevitable problem
but does the job better in more serious problems.

I am happy with Mathematica!!!

Dimitris

 /  dimitris       :
> Hi fellas.
> This appeared in another forum as part of a question
> what another CAS does.
> Just of curiosity I check Mathematica's performance (5.2).
> The result was poor!
>
> Here is the expression
>
>
> In[16]:=
> o = (Log[2]*Cos[Pi/12] - Log[2]*Sin[Pi/12] - 2*Cos[Pi/12] + 2*Sin[Pi/
> 12] + Sqrt[2] +
>     2*Log[Cos[Pi/12] - Sin[Pi/12]]*Cos[Pi/12] - 2*Log[Cos[Pi/12] -
> Sin[Pi/12]]*Sin[Pi/12])/
>    (Log[2]*Cos[Pi/12] - Log[2]*Sin[Pi/12] + 2*Log[Cos[Pi/12] -
> Sin[Pi/
> 12]]*Cos[Pi/12] -
>     2*Log[Cos[Pi/12] - Sin[Pi/12]]*Sin[Pi/12])
>
>
> Out[16]=
> (Sqrt[2] + (-1 + Sqrt[3])/Sqrt[2] - (1 + Sqrt[3])/Sqrt[2] - ((-1 +
> Sqrt[3])*Log[2])/(2*Sqrt[2]) +
>    ((1 + Sqrt[3])*Log[2])/(2*Sqrt[2]) - ((-1 + Sqrt[3])*Log[-((-1 +
> Sqrt[3])/(2*Sqrt[2])) + (1 + Sqrt[3])/(2*Sqrt[2])])/
>     Sqrt[2] + ((1 + Sqrt[3])*Log[-((-1 + Sqrt[3])/(2*Sqrt[2])) + (1 +
> Sqrt[3])/(2*Sqrt[2])])/Sqrt[2])/
>   (-(((-1 + Sqrt[3])*Log[2])/(2*Sqrt[2])) + ((1 + Sqrt[3])*Log[2])/
> (2*Sqrt[2]) -
>    ((-1 + Sqrt[3])*Log[-((-1 + Sqrt[3])/(2*Sqrt[2])) + (1 + Sqrt[3])/
> (2*Sqrt[2])])/Sqrt[2] +
>    ((1 + Sqrt[3])*Log[-((-1 + Sqrt[3])/(2*Sqrt[2])) + (1 + Sqrt[3])/
> (2*Sqrt[2])])/Sqrt[2])
>
> Watch now a really bad performance!
>
> In[17]:=
> (Simplify[#1[o]] & ) /@ {Numerator, Denominator}
>
> Out[17]=
> {0, 0}
>
> That is Mathematica simplifies succesfully both the numerator
> and denominator to zero. So, you wonder what goes wrong?
>
> Try now to simplify the whole expression!
>
> In[19]:=
> Simplify[o]
>
> Out[19]=
> 1
>
> A very weird result to my opinion!
> Simplification of 0/0 to 1?
> I think no simplification or some
> warning messages would be much better
> than 1!
>
> Note also that
>
> In[20]:=
> RootReduce[o]
>
> Out[20]=
> 1
>
> Dimitris



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